Question: Simplify and expand the following expression: $ \dfrac{3}{t + 10}+ \dfrac{1}{t - 5}- \dfrac{3}{t^2 + 5t - 50} $
Explanation: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor the quadratic in the third term: $ \dfrac{3}{t^2 + 5t - 50} = \dfrac{3}{(t + 10)(t - 5)}$ Now we have: $ \dfrac{3}{t + 10}+ \dfrac{1}{t - 5}- \dfrac{3}{(t + 10)(t - 5)} $ The least common multiple of the denominators is: $ (t + 10)(t - 5)$ In order to get the first term over $(t + 10)(t - 5)$ , multiply by $\dfrac{t - 5}{t - 5}$ $ \dfrac{3}{t + 10} \times \dfrac{t - 5}{t - 5} = \dfrac{3(t - 5)}{(t + 10)(t - 5)} $ In order to get the second term over $(t + 10)(t - 5)$ , multiply by $\dfrac{t + 10}{t + 10}$ $ \dfrac{1}{t - 5} \times \dfrac{t + 10}{t + 10} = \dfrac{t + 10}{(t + 10)(t - 5)} $ Now we have: $ \dfrac{3(t - 5)}{(t + 10)(t - 5)} + \dfrac{t + 10}{(t + 10)(t - 5)} - \dfrac{3}{(t + 10)(t - 5)} $ $ = \dfrac{ 3(t - 5) + t + 10 - 3} {(t + 10)(t - 5)} $ Expand: $ = \dfrac{3t - 15 + t + 10 - 3}{t^2 + 5t - 50} $ $ = \dfrac{4t - 8}{t^2 + 5t - 50}$